Find the integrating factor of the differential equation:
`(dx)/(dy) = (y tan y -xtan y - xy )/(y tan y)`

Answer

To find the integrating factor of the given differential equation:

\[ \frac{dx}{dy} = \frac{y \tan y - x \tan y - xy}{y \tan y} \]

we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the equation in a more manageable form. We can simplify the right-hand side:

\[ \frac{dx}{dy} = \frac{y \tan y - x \tan y - xy}{y \tan y} \]

This can be rewritten as:

\[ \frac{dx}{dy} = \frac{y \tan y}{y \tan y} - \frac{x \tan y}{y \tan y} - \frac{xy}{y \tan y} \]

This simplifies to:

\[ \frac{dx}{dy} = 1 - \frac{x}{y} - \frac{x}{\tan y} \]

### Step 2: Rearrange the Equation
Now we can rearrange the equation to isolate \(x\):

\[ \frac{dx}{dy} + \left(\frac{1}{y} + \frac{1}{\tan y}\right)x = 1 \]

This is now in the standard linear form:

\[ \frac{dx}{dy} + P(y)x = Q(y) \]

where \(P(y) = \frac{1}{y} + \frac{1}{\tan y}\) and \(Q(y) = 1\).

### Step 3: Find the Integrating Factor
The integrating factor \(\mu(y)\) is given by:

\[ \mu(y) = e^{\int P(y) \, dy} \]

Calculating \(P(y)\):

\[ P(y) = \frac{1}{y} + \cot y \]

Now we find the integral:

\[ \int P(y) \, dy = \int \left(\frac{1}{y} + \cot y\right) \, dy \]

This can be split into two integrals:

\[ \int \frac{1}{y} \, dy + \int \cot y \, dy \]

### Step 4: Evaluate the Integrals
We know:

\[ \int \frac{1}{y} \, dy = \log |y| \]

and

\[ \int \cot y \, dy = \log |\sin y| \]

Thus, combining these results:

\[ \int P(y) \, dy = \log |y| + \log |\sin y| = \log |y \sin y| \]

### Step 5: Calculate the Integrating Factor
Now we can find the integrating factor:

\[ \mu(y) = e^{\log |y \sin y|} = |y \sin y| \]

Since we are typically interested in the positive value, we can write:

\[ \mu(y) = y \sin y \]

### Final Result
The integrating factor of the given differential equation is:

\[ \mu(y) = y \sin y \]